y^'=(y-1)^{1/3}

Lösung

y^{^{'}} = (y - 1)^{\frac{1}{3}}

y = \frac{( 8 t ^{3} + 8 c _{1}^{3} + 3 ( 8 c _{1} t ^{2} + 8 c _{1}^{2} t ) ) ^{\frac{1}{2}}}{3 \cdot 3 ^{\frac{1}{2}}} + 1, y = - \frac{( 8 t ^{3} + 8 c _{1}^{3} + 3 ( 8 c _{1} t ^{2} + 8 c _{1}^{2} t ) ) ^{\frac{1}{2}}}{3 \cdot 3 ^{\frac{1}{2}}} + 1

Schritte zur Lösung

y^{'} (t) = (y - 1)^{\frac{1}{3}}

Solve separable ODE:

y = \frac{( 8 t ^{3} + 8 c _{1}^{3} + 3 ( 8 c _{1} t ^{2} + 8 c _{1}^{2} t ) ) ^{\frac{1}{2}}}{3 \cdot 3 ^{\frac{1}{2}}} + 1, y = - \frac{( 8 t ^{3} + 8 c _{1}^{3} + 3 ( 8 c _{1} t ^{2} + 8 c _{1}^{2} t ) ) ^{\frac{1}{2}}}{3 \cdot 3 ^{\frac{1}{2}}} + 1

y = \frac{( 8 t ^{3} + 8 c _{1}^{3} + 3 ( 8 c _{1} t ^{2} + 8 c _{1}^{2} t ) ) ^{\frac{1}{2}}}{3 \cdot 3 ^{\frac{1}{2}}} + 1, y = - \frac{( 8 t ^{3} + 8 c _{1}^{3} + 3 ( 8 c _{1} t ^{2} + 8 c _{1}^{2} t ) ) ^{\frac{1}{2}}}{3 \cdot 3 ^{\frac{1}{2}}} + 1

y y^{^{'}} - x = 0

\frac{d r}{d t} = - 4 r t, r = r^{0}, t = 0

y^{^{'}} = (3 + sin (t)) y^{2}

y^{^{'}} = - x^{2} \cdot (y)^{2}

\frac{d y}{d x} = \frac{- 0.7}{cos ( y )}, y (0) = \frac{π}{3}